How many bit strings of length 10 have exactly three 0's, more 0's than 1's, at least seven 1's, at least three 1's?
Any help would be appreciated.
You need to choose three positions, from the ten, in which to place the zeros. So there are $\displaystyle \binom{10}3$ possibilities.
That means there are at least six 0's. Such strings can have exactly six 0's, exactly seven 0's, exactly eight 0's, exactly nine 0's, or exactly ten 0's. So the total number of combinations ismore 0's than 1's
$\displaystyle \binom{10}6+\binom{10}7+\binom{10}8+\binom{10}9+\b inom{10}{10}\text.$
This is similar to the above.at least seven 1's
You could solve it like the above problems, but it is easier to find the number of bit strings having less than three 1's, and subtracting that from the number of all possible bit strings of length 10.at least three 1's?